Question

An athlete on the training ground starts at point and runs 120 m South to point B, then runs 200 m East to point C and the runs 270 m North to point D. The points on the training ground are shown in the diagram below: 2.2. 2.3 B Use the tail to head method, draw a neat, fully labelled displacement vector diagram and include the resultant displacement of the athlete. D Is the vector diagram in question 2.2, closed or not closed vector diagram? Explain the answer. 2.4 Calculate the 2.4.1 magnitude of the resultant displacement of the athlete 2.4.2 direction of the resultant displacement (4) (2) (4) 6 [15]​

Answer 1

To draw the displacement vector diagram, we start at point A and draw a vector from A to B, representing the athlete's displacement of 120 m South. We then draw a vector from the end of the first vector (B) to the end of the second vector (C), representing the athlete's displacement of 200 m East. Finally, we draw a vector from the end of the second vector (C) to point D, representing the athlete's displacement of 270 m North. The diagram should form a closed triangle.

To find the resultant displacement of the athlete, we can use the Pythagorean theorem and trigonometry. Let's call the displacement from A to B "vector AB," the displacement from B to C "vector BC," and the displacement from C to D "vector CD." The magnitude of the resultant displacement (R) is given by:

R = √(AB² + BC² + CD²)

R = √(120² + 200² + 270²) = 334.4 m (rounded to one decimal place)

To find the direction of the resultant displacement, we can use trigonometry. We can find the angle between the resultant displacement and the North direction using the following formula:

θ = tan⁻¹[(ΣN)/(ΣE)]

Where ΣN is the sum of the Northward components of the displacement vectors (in this case, CD), and ΣE is the sum of the Eastward components of the displacement vectors (in this case, BC).

θ = tan⁻¹[(270 m)/(200 m)] = 53.1° (rounded to one decimal place)

Therefore, the magnitude of the resultant displacement is 334.4 m and the direction of the resultant displacement is 53.1° North of East.